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STRUCTURE OF CARBON ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism " (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Carbon © has 15 known isotopes, from C-8 to C-22, 2 of which (C-12 and C-13) are stable. The longest-lived radioisotope is C-14, with a half-life of 5,700 years. This is also the only carbon radioisotope found in nature - trace quantities are formed cosmogenically by the reaction N-14 + 1n → C-14 + 1H. In order to reveal the structure of carbon isotopes I start with the diagram of my paper STRUCTURE OF C-12, C-14 AND N-14 . In this diagram you see that the nucleons of the first horizontal plane have positive spins (+HP1), while the nucleons of the second horizontal plane have negative spins (-HP2). In the absence of 4 neutrons we get the C-8. Since the C-8 has S =0 we conclude that it has only the neutrons n3(+1/2) and n4(-1/2) . Since these neutrons exist in both horizontal planes of opposite spins they are able to give bonds with all protons. DIAGRAM OF C-12 p2(-1/2)..........n4 (-1/2)……….p6(-1/2) n2 (-1/2)..........p4 (- 1/2)……….n6(-1/2) -HP2 n1(+1/2)..........p3(+1/2)……….n5(+1/2) p1(+1/2)..........n3(+1/2)……….p5(+1/2) +HP1 However in the absence of 3 neutrons we get the C-9 with S= -3/2, which means that it has a structure not of two horizontal planes but of three ones, in which the first horizontal plane has 3 nucleons of negative spins. In other words it has a structure similar o the structure of O-16 . (See my STRUCTURE OF O-16, O-15 ). Especially the structure of this nuclide in the first horizontal plane (-HP1) has two protons and one neutron of negative spins. It gives the total spin S =-3/2 , because the second plane (+HP2) and the third plane ( -HP3) of opposite spins have 4 protons and 2 neutrons giving S =0. On the other hand in the absence of two neutrons we get the C-10 of S=0 which has the structure of the diagram of C-12 in which the neutrons n1 and n6 of opposite spins are absent. However in the absence of n1(+1/2) we get the C-11 which has not S =-1/2 but S =-3/2. It means that the n5(+1/2) changes the spin from S =+1/2 to S=-1/2 because it is moved from the +HP1 to -HP2 in order to make a single horizontal bond with p6(-1/2). It is not shown because it exists behind the p6. Of course the C-12 has S=0 because it has 3 protons and 3 neutrons in each horizontal plane. For understanding the stable C-13 of S =-1/2 with the extra neutron n7(-1/2) we conclude that it makes a single bond with the p4(-1/2). It is not shown in the diagram because it is in front of p4(-1/2). Since the p4(-1/2) makes 5 bonds with n7, n6, n4, n3 and n2, one concludes that they contribute to the increase of the binding energy of the single bond (p4n7) which leads to the stability of C-13. However the C-14 of S=0 is unstable because the extra n8(+1/2) which makes the n8p3 bond behind the p3 exerts a new repulsive force with the n7. In other words the new n8n7 repulsion of short range leads to the beta decay. In the C-15 of S =+1/2 the third extra n9(+1/2) makes the single bond n9p1 which is in front of p1. In this case the new n9n8 and n9n7 repulsions of short range lead to the beta decay. In the C-16 of S=0 the four extra neutrons of opposite spins like n7(+1/2), n8(-1/2), n9(+1/2 )and n10(-1/2) make single horizontal bonds with p3(+1/2), p4(-1/2), p1(+1/2) and p6(-1/2) respectively. They are not shown because they exist behind the protons and in front of them. However in the C-17 of S= +3/2 the fifth extra neutron like the n11(+1/2) makes a single horizontal bond with p5(+1/2) existing in front of the p5. Whereas the n10(-1/2) changes the spin from S =-1/2 to S=+1/2 because it is moved from the p6 to p5 in order to make a single horizontal bond with p5 existing on the right side of the parallelepiped. Since this change gives S =+1 one concludes that the total spin S =+3/2 is due to the summation of S =+1 and the S =+1/2 of the additional n11(+1/2). Now for understanding the structure of C-18 with S=0 we conclude that the 6 extra neutrons of opposite spins make 6 extra single bonds with the 6 protons of opposite spins. They are not shown because they exist in front of protons and behind them. Then for revealing the structure of C-19 with S =+1/2 we suggest that the seventh extra neutron of spin S =+1/2 makes a single bond with p1 existing on the left side of the parallelepiped. Similarly for understanding the structures of C-20 with S=0 ,the C-21 with S=+1/2 and the C-22 with S=0 we suggest that the three more neutrons make single bonds with p2(-1/2), the p5(+1/2) and the p6(-1/2) respectively. The bonds are not shown because they exist on the left and on the right sides of the parallelepiped.. Category:Fundamental physics concepts